1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integrals involving trig functions

  1. Jul 17, 2010 #1
    1. The problem statement, all variables and given/known data

    I need help evaluating the integral Cotx3/10

    I factored out the 1/10 from the integral and am just left with (1/10)*Cotx3

    from here i do not really know what to do. I rewrote it in terms of sine and cosine to get


    I multiply the integral by (1/Sinx3) to get rid of the denominator and am left with


    I factor out a (Cosx2) and am left with


    Rewriting using trig indentities, I get


    rewriting i get


    I solve this integral and get

    (1/10)sinx - (1/10)*((sinx)3/3) + C

    but that is incorrect. The answer is supposed to be


    What went wrong?
  2. jcsd
  3. Jul 17, 2010 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Well, for one thing, you can't just get rid of parts of the integrand.
    Just to be clear, you are trying to compute

    [tex]\int \cot^3 x\,dx[/tex]

    and not

    [tex]\int \cot x^3\, dx[/tex]


    Hint: Try the substitution u=sin x.
  4. Jul 17, 2010 #3
    Yes, Cot3x
  5. Jul 18, 2010 #4


    User Avatar
    Science Advisor

    And is it
    [tex]\frac{cot^3 (x)}{10}[/tex]

    Try writing [itex]cot^3(x)[/itex] as

    [tex]\frac{sin^3(x)}{cos^3(x)}= [/tex][tex]\frac{sin^2(x)}{cos^3(x)}cos(x)=[/tex][tex] \frac{1- cos^2(x)}{cos^3(x)} sin(x)[/tex]

    and use the substitution u= cos(x).
  6. Jul 18, 2010 #5
    It is [tex]\frac{cot^3 (x)}{10}[/tex]

    I actually have no idea how to use most of the math tags, so writing it out is hard.

    Cot3x is [tex]\frac{Cos^3(x)}{Sin^3(x)}[/tex]

    I'll try distributing out a cosine factor and converting the rest to sign and see how it goes
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook